Geometry circle problem and the curves $y= \pm \frac{1}{x}$ 
A circle touches the $y-$axis at the origin and the curves $y= \frac{1}{x}$,  $y= -\frac{1}{x}$ when $x>0$. Determine the radius of the circle.


I managed to do the question, but it got pretty messy and terrible. By first denoting the center as $C(x_0, y_0)$ we can deduce that the radius is simply $x_0$. Then defining the points $A(a, \frac{1}{a})$ and $B(b, -\frac{1}{b})$ and drawing the tangents from the center to the points where the curves touch the circle I could algebraicly solve this and here it went messy. I had to use the derivative and the property that the product of two gradients equal $-1$ in order to first find point $a$ from what I could use to find $x_0$. Is there some clean way to do this or is it just one of those problems where it gets messy anyway?
 A: Not a proper answer (yet?), but an illustration that reveals the interesting geometry of the situation:

Here, $|OA|=1$, $|OB|=\sqrt{3}$ (which is easily constructible from $\overline{OA}$), $|OC|=\sqrt[4]{3}$ (constructed as the geometric mean of $|OA|$ and $|OB|$). Then $|OP|=\tfrac23|OC|$, and $\triangle PQR$ is equilateral.
(I omitted the semicircle on $\overline{AB}$ that yields $C$, since that semicircle almost passes through $Q$, which could cause confusion.)
A: Let $c>0$ denote the center of the circle. Equating and squaring, we have
$$
(\pm x^{-1})^2 = c^2 - (x-c)^2
$$
$$
x^{-2} = 2cx-x^2
$$
$$
x^4-2cx^3+1=0
$$The discriminant of this equation is $256-432 c^4$. But since the curves must lie tangent, this discriminant must be $0$. If it's negative, there will be a double root. If it's positive, the curves won't intersect. Solving for $c$, we have $c=2\cdot 3^{-3/4}\approx 0.877$.
A: It is quite easy. By the symmetry of the problem, we know the circle's center must lie along the $x$-axis. So we only need one coordinate to describe it, which we will call $c$. Then the circle must have equation
$$(x - c)^2 + y^2 = r^2$$
Moreover, since the left hand extremity of the circle should touch the $y$-axis, then a line from the center to that touching point is a radius, so $r = c$. Thus
$$(x - c)^2 + y^2 = c^2$$
and we can expand out to get
$$\begin{align}
(x^2 - 2cx + c^2) + y^2 &= c^2\\
x^2 + y^2 - 2cx + c^2 &= c^2\\
x^2 + y^2 &= 2cx
\end{align}$$
So we now have a growable circle coming off the y-axis, that we can let the little thing grow up (by letting $c$ do so) just as much as need be so he juuust bumps/kisses the curve (❤︎#mehhr). Hence, we must find the limiting case where only a single point of the circle lies upon the curve. So take $y = \pm \frac{1}{x} = \pm x^{-1}$ in the above, to get
$$x^2 + x^{-2} = 2cx$$
Times $x^2$, you get
$$x^4 + 1 = 2cx^3$$
so we want to find which $c$ makes
$$x^4 - 2cx^3 + 1 = 0$$
have only a single solution (coordinate of point of intersection) in $x$. This is most easily done with calculus, though there are purely algebraic methods but a bit nontrivial since this is a quartic (degree 4) polynomial equation. Differentiating, we get that the minimum point is when $4x^3 - 6cx^2 = 0$ so $4x - 6c = 0$ and $x = \frac{6c}{4} = \frac{3c}{2}$. The height of the minimum is then
$$\begin{align}
\left(\frac{3c}{2}\right)^4 - 2c\left(\frac{3c}{2}\right)^3 + 1 &= \frac{81c^4}{16} - 2c\left(\frac{27c^3}{8}\right) + 1\\
 &= \frac{81c^4}{16} - \frac{54c^4}{8} + 1\\
 &= \frac{81c^4}{16} - \frac{108c^4}{16} + \frac{16}{16}\\
 &= \frac{-27c^4 + 16}{16}\end{align}$$
So this height is 0 when $16 = 27c^4$ or $c = r = \sqrt[4]{\frac{16}{27}}$.
Thus the answer is the circle at $\left(\sqrt[4]{\frac{16}{27}}, 0\right)$ and with radius $\sqrt[4]{\frac{16}{27}}$. For scale, this center and radius are approximately 0.877 units, which looks to be about right from your graph given the center is just a bit to the left of 1 on the x-axis. This number seems to be what everybody else is getting, just expressed in different ways.
A: The equation of a circle by the origin and of center $(0,a)$ is $$(x-a)^2+y^2=a^2.$$ 
We express that it intersects the hyperbola,
$$(x-a)^2+\frac1{x^2}=a^2,$$ forming a double root.
Hence,
$$x^4-2ax^3+1=4x^3-6ax^2=0.$$
Finally,
$$a=\frac23x=\frac23\sqrt[4]3.$$
A: I used a calculus based approach as well. It's not messy if you keep it simple.
First deduce the form of the circle. The circle will be symmetrical about the $x$ axis Start by defining the centre $(a,0)$. The general equation will be $(x-a)^2 + y^2 = r^2$, and by setting $x=y=0$, you can immediately deduce $r = a$.
Now let the point the upper half of the circle touches the upper part of the hyperbola have the $x$ coordinate $b$.
By commonality of the points, you have $\displaystyle (b-a)^2 + y_b^2 = a^2 \implies y_b = \sqrt{b(2a-b)}$. Since this point also lies on the hyperbola, you can form the equation $\displaystyle \sqrt{b(2a-b)} = \frac 1b$ which can be squared and rearranged easily to $\displaystyle 2a-b = \frac 1{b^3}$. Call this equation $1$.
Next consider the slopes of the tangents at that point of "touching". By implicit differentiation of the equation for the circle, you get $\displaystyle 2(x-a) + 2yy' = 0$ so at $x = b$, the slope is given by $\displaystyle \frac{a-b}{\sqrt{b(2a-b)}}$. This will be equal to the slope of the hyperbola, i.e. $\displaystyle -\frac 1{b^2}$. Using the result $\displaystyle \sqrt{b(2a-b)} = \frac 1b$ from above, we get $\displaystyle b(a-b) = - \frac 1{b^2} \implies b-a = \frac 1{b^3}$. Call this equation $2$.
Comparing equations $1$ and $2$, we get: $\displaystyle 2a-b = b-a \implies b  = \frac 32 a$. Substitution into either equation would quickly allow you to solve for $a$. Let's use equation $2$: $\displaystyle \frac 32 a - a = (\frac 8{27})\frac 1 {a^3} \implies a^4 = \frac{16}{27} \implies a = \frac 2{27^{\frac 14}}= \frac 23 \sqrt[4] 3$. 
So the radius of the circle is $\displaystyle \frac 23 \sqrt[4] 3$.
(And that's also the $x$ coordinate of the centre. And to get the $x$ coordinate of the touching point with the hyperbola, just multiply that by $\displaystyle \frac 32$ to get $\displaystyle \sqrt[4] 3$).
 Desmos link
A: Let $(a,\frac1a)$ be the touch point on $y=\frac1x$. Then, match the radius $r$ as well as the tangent $y’=-\frac1{a^2}$ to establish
$$(a-r)^2+\frac1{a^2}=r^2,\>\>\>\>\> \frac{\frac1a-0}{a-r} = -\frac1{y’} =a^2$$
Simplify 
$$a^4+1=2ra^3,\>\>\>\>\> a^4-1=ra^3$$
Then, eliminate $ra^3$ to obtain $a=3^{1/4}$ and the radius $r= \frac2{3^{3/4}}$.
